Is ${49667}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {49667}= &&{4}\cdot10000+ \\&&{9}\cdot1000+ \\&&{6}\cdot100+ \\&&{6}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {49667}= &&{4}(9999+1)+ \\&&{9}(999+1)+ \\&&{6}(99+1)+ \\&&{6}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {49667}= &&\gray{4\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {4}+{9}+{6}+{6}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${49667}$ is divisible by $9$ if ${ 4}+{9}+{6}+{6}+{7}$ is divisible by $9$ Add the digits of ${49667}$ $ {4}+{9}+{6}+{6}+{7} = {32} $ If ${32}$ is divisible by $9$ , then ${49667}$ must also be divisible by $9$ ${32}$ is not divisible by $9$, therefore ${49667}$ must not be divisible by $9$.